Consider the matrix $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$.
Given below are two statements :
Statement I : $ f(-x)$ is the inverse of the matrix $f(x)$.
Statement II : $f(x) f(y)=f(x+y)$.
In the light of the above statements, choose the correct answer from the options given below :
Solution
<p>$$\begin{aligned}
& f(-x)=\left[\begin{array}{ccc}
\cos x & \sin x & 0 \\
-\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right] \\
& f(x) \cdot f(-x)=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=I
\end{aligned}$$</p>
<p>Hence statement- I is correct</p>
<p>Now, checking statement II</p>
<p>$$\begin{aligned}
& f(y)=\left[\begin{array}{ccc}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{array}\right] \\
& f(x) \cdot f(y)=\left[\begin{array}{ccc}
\cos (x+y) & -\sin (x+y) & 0 \\
\sin (x+y) & \cos (x+y) & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \Rightarrow f(x) \cdot f(y)=f(x+y)
\end{aligned}$$</p>
<p>Hence statement-II is also correct.</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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